Problem: Solve for $x$ : $ 5|x - 3| - 5 = -5|x - 3| + 2 $
Answer: Add $ {5|x - 3|} $ to both sides: $ \begin{eqnarray} 5|x - 3| - 5 &=& -5|x - 3| + 2 \\ \\ { + 5|x - 3|} && { + 5|x - 3|} \\ \\ 10|x - 3| - 5 &=& 2 \end{eqnarray} $ Add ${5}$ to both sides: $ \begin{eqnarray} 10|x - 3| - 5 &=& 2 \\ \\ { + 5} &=& { + 5} \\ \\ 10|x - 3| &=& 7 \end{eqnarray} $ Divide both sides by ${10}$ $ \dfrac{10|x - 3|} {{10}} = \dfrac{7} {{10}} $ Simplify: $ |x - 3| = \dfrac{7}{10}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 3 = -\dfrac{7}{10} $ or $ x - 3 = \dfrac{7}{10} $ Solve for the solution where $x - 3$ is negative: $ x - 3 = -\dfrac{7}{10} $ Add ${3}$ to both sides: $ \begin{eqnarray} x - 3 &=& -\dfrac{7}{10} \\ \\ {+ 3} && {+ 3} \\ \\ x &=& -\dfrac{7}{10} + 3 \end{eqnarray} $ Change the ${ + 3}$ to an equivalent fraction with a denominator of $10$ $ x = - \dfrac{7}{10} {+ \dfrac{30}{10}} $ $ x = \dfrac{23}{10} $ Then calculate the solution where $x - 3$ is positive: $ x - 3 = \dfrac{7}{10} $ Add ${3}$ to both sides: $ \begin{eqnarray} x - 3 &=& \dfrac{7}{10} \\ \\ {+ 3} && {+ 3} \\ \\ x &=& \dfrac{7}{10} + 3 \end{eqnarray} $ Change the ${ + 3}$ to an equivalent fraction with a denominator of $10$ $ x = \dfrac{7}{10} {+ \dfrac{30}{10}} $ $ x = \dfrac{37}{10} $ Thus, the correct answer is $x = \dfrac{23}{10} $ or $x = \dfrac{37}{10} $.